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In Exercises \(11-22,\) evaluate the iterated integral. $$ \int_{0}^{1} \int_{0}^{2}(x+y) d y d x $$

Short Answer

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Step by step solution

01

Evaluate Inner Integral

First, evaluate the inner integral with respect to y: \(\int_{0}^{2}(x+y) dy\). Treat x as a constant and integrate with y as the variable. The antiderivative of y is \(\frac{1}{2}y^2\), and the antiderivative of x (considered as a constant) is \(xy\). Including the limits, the result of the integral would be \([x*2 + \frac{1}{2}*2^2] - [x*0 + \frac{1}{2}*0^2] = 2x + 2\).
02

Evaluate Outer Integral

Next, evaluate the outer integral with respect to x: \(\int_{0}^{1}(2x + 2) dx\). The antiderivative of 2x is \(x^2\) and the antiderivative of 2 is \(2x\). Including the limits, the result of the integral would be \([1^2 + 2*1] - [0^2 + 2*0] = 1 + 2 = 3\).
03

Final Result

Thus the volume under the function f(x,y) = (x + y) over the given region in the xy-plane is 3 units^3.

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